1. 31 January, 2 February, 2005 Chapter 6 Genetic Recombination in Eukaryotes Linkage and genetic diversity

2. Overview In meiosis, recombinant products with new combinations of parental alleles are generated by: –independent assortment (segregation) of alleles on nonhomologous chromosomes. –crossing-over in premeiotic S between nonsister homologs. In dihybrid meiosis, 50% recombinants indicates either that genes are on different chromosomes or that they are far apart on the same chromosome. Recombination frequencies can be used to map gene loci to relative positions; such maps are linear. Crossing-over involves formation of DNA heteroduplex.

3. Overview

4. Recombination Detailed

5. Independent Assortment

6. Independent assortment (2) For genes on different (nonhomologous) pairs of chromosomes, recombinant frequency is always 50% A/A ; B/B  a/a ; b/b A/a ; B/b ¼ A ; B P ¼ A ; bR ¼ a ; BR ¼ a ; bP A/A ; b/b  a/a ; B/B A/a ; B/b ¼ A ; B R ¼ A ; bP ¼ a ; BP ¼ a ; bR 50% recombinants

7. Testcross of a Dihybrid

8. Self Cross of a Dihybrid

11. Independent assortment: multiple loci Calculations can be made for any gene combination using predicted outcomes at single loci and the product rule P 1 A/a ; B/b ; C/c ; D/d  P 2 a/a ; B/b ; C/c ; D/D # gametes P 1 2 x 2 x 2 x 2 = 16 # gametes P 2 1 x 2 x 2 x 1 = 4 # genotypes in F 1 2 x 3 x 3 x 2 = 36 # phenotypes in F 1 2 x 2 x 2 x 1 = 8 Frequency of A/– ; B/– ; C/– ; D/– ½ x ¾ x ¾ x 1 = 9/32

12. Crossing-over Breakage and rejoining of homologous DNA double helices Occurs only between nonsister chromatids at the same precise place Visible in diplotene as chiasmata Occurs between linked loci on same chromosome –cis: recessive alleles on same homolog ( AB/ab ) –trans: recessive alleles on different homologs ( Ab/aB )

17. Linkage maps RF is (60+50)/400=27.5%, clearly less than 50% Map is given by: # observed 140 50 60 150 AB 27.5 m.u.

20. Trihybrid testcross Sometimes called three-point testcross Determines gene order as well as relative gene distances 8 categories of offspring –for linked genes, significant departure from 1:1:1:1:1:1:1:1 Works best with large numbers of offspring, as in fungi, Drosophila

21. Trihybrid testcross example v (vestigial) v + (long) wings b (black) b + (gray) body p (purple) p + (red) eyes v + v +. b b. pp x vv. b + b +. p + p + v/ v +. b/ b +. p/ p +

22. v/ v +. b/ b +. p/ p + xv/v.b/b.p/p Progeny: vb + p + 580Parental v + bp592Parental vbp + 45 v + b + p 40 vbp 89 v + b + p + 94 vb + p 3 v + bp + 5

23. v.b and v +.b + 45 + 40 + 89 + 94 = 268 / 1448 =.185 v.p and v +.p + 89 + 94 + 3 + 5 = 191 / 1448 =.132 b.p + and b +.p 45 + 40 + 3 + 5 = 93 / 1448 =.064 v 13.2 p6.4 b Why do 13.2 + 6.4 = 19.6 instead of 18.5?

24. The failure to count double crossover gametes as v.b recombinants results in an underestimation of the v.b distance. A better v.b number would be 268 + 3 + 3 + 5 + 5 = 284 / 1448 = 19.6. Aaah! In general, to minimize the effect of double crossovers, it is necessary to measure a number of small RF distances and sum to larger distances.

25. Interference Crossing-over in one region of chromosome sometimes influences crossing-over in an adjacent region Interference = 1 – (coefficient of coincidence) Usually, I varies from 0 to 1, but sometimes it is negative, meaning double crossing-over is enhanced

26. In the previous example, expect (.132 x.064 =.0084)(1448) = 12 double crossovers, but saw 8, so I = 1-(8/12) = 4/12 = 1/3. In regions where double crossovers are forbidden, observed = 0, so I = 1. What do you think negative interference means?

27. Genetic maps Useful in understanding and experimenting with the genome of organisms Available for many organisms in the literature and at Web sites Maps based on RF are supplemented with maps based on molecular markers, segments of chromosomes with different nucleotide sequences

29. Chi-square test Statistical analysis of goodness of fit between observed data and expected outcome (null hypothesis) Calculates the probability of chance deviations from expectation if hypothesis is true 5% cutoff for rejecting hypothesis –may therefore reject true hypothesis –statistical tests never provide certainty, merely probability

30. Chi-square application to linkage Null hypothesis for linkage analysis –based on independent assortment, i.e., no linkage –no precise prediction for linked genes in absence of map for all classes Calculated from actual observed (O) and expected (E) numbers, not percentages

31. What is E? For the AbBb testcross example from the text, E is not just 125 for each of the genotypes, as there may be allele effects on viability. Instead, get expected values from the data: A a B142112254 b113133246 255245 So, E (AB) would be (255/500) x (254/500) x 500 = 129.54 Other E are found similarly, and Chi-square is 4.97

33. Mechanism of meiotic crossing-over Exact mechanism with no gain or loss of genetic material Current model: heteroduplex DNA –hybrid DNA molecule of single strand from each of two nonsister chromatids –heteroduplex resolved by DNA repair mechanisms May result in aberrant ratios in systems that allow their detection

36. Recombination within a gene Recombination between alleles at a single locus In diploid heterozygous for mutant alleles of the same gene, recombination can generate wild-type and double mutant alleles Rare event, 10 -3 to 10 -6, but in systems with large number of offspring, recombination can be used to map mutations within a gene a 1 /a 2  a + and a 1,2

37. Assignment: Concept map, solved problems 1 and 2, all other basic and challenging problems Continue with PubMed section of the Web tutorial.